The chapter's decision section. Sections 7.1 and 7.2 supplied the distributions of a sample mean and of a sum; this one supplies the question that must be answered before either can be used, namely whether the problem concerns an individual value, a mean, or a total. The book is explicit that a question about an individual value must be answered from that variable's own distribution and not from the central limit theorem at all. The section also states the law of large numbers, which says that as samples get larger their means get closer to the population mean, and observes that the central limit theorem illustrates it because the standard error shrinks toward zero. Two worked examples take populations that are conspicuously not normal, a uniform and an exponential, which is where the theorem does real work rather than restating what chapter 6 already covered.
Subject: Statistics · 65 slides · symbolic lesson
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Title
Statistics · Chapter 7 — The Central Limit Theorem
Using the Central Limit Theorem
Objectives
Five outcomes, and the first has to be settled before any of the others can be attempted.
OpenStax Introductory Statistics 2e, §7.3 Using the Central Limit Theorem §7.3, pp. 373-380 — the section these objectives are drawn from
Warm-up
Sections 7.1 and 7.2 gave two distributions built from the same population; chapters 4 and 5 gave the population's own.
Discussion prompt
Excess mobile minutes follow an exponential distribution with mean 22. What is the probability that one customer exceeds 20 minutes, and what is the probability that the mean of 80 customers does?
Hint: These are not the same question, and the answers are not close.
Answer:
For one customer the exponential applies directly: the right tail is e to the minus twenty over twenty-two, about 0.4029 — a bit under half.
For a mean of 80 the central limit theorem applies, giving a normal distribution centred at 22 with a standard error of 22 over the square root of 80, about 2.46. Twenty is then 0.81 standard errors BELOW the centre, so the probability of exceeding it is about 0.7919 — nearly double.
Both are correct, and they differ because they describe different quantities. The book makes this the central lesson of the section, and its rule is blunt: when asked about an individual value, do not use the theorem.
Concept
It is important to understand when to use the central limit theorem. If you are being asked to find the probability of the mean, use it for the mean. If you are being asked to find the probability of a sum or total, use it for sums. This also applies to percentiles. If you are being asked to find the probability of an individual value, do not use the theorem — use the distribution of its random variable.
choosing the distribution — The first step of every problem in this chapter. One population supplies three distributions — its own, its sample mean's and its sum's — and only the wording of the question determines which applies.
\[ X, \qquad \bar{X} \sim N\left(\mu, \tfrac{\sigma}{\sqrt{n}}\right), \qquad \sum X \sim N\left(n\mu, \sqrt{n}\,\sigma\right) \]
The warning about individual values is worth stating as its own rule because it is the one error the theorem itself invites. Having spent two sections learning that things become normal, it is tempting to treat everything as normal — but the theorem says nothing whatever about individual values, and applying it to one replaces a correct distribution with a wrong one.
Figure (svg): A decision procedure for choosing between the population, the mean and the sum
OpenStax Introductory Statistics 2e, §7.3 Using the Central Limit Theorem §7.3, pp. 373-374
Section
Section 1
Concept
The same population supports three different distributions, and the question's wording decides which is wanted. A question about one observation uses the population's own distribution; one about an average uses the sampling distribution of the mean; one about a total uses the distribution of the sum.
the individual-value rule — If the question asks about a single observation, the central limit theorem does not apply. Use the stated distribution of the random variable, whatever shape it has.
\[ \text{one value} \to X; \quad \text{average} \to \bar{X}; \quad \text{total} \to \sum X \]
The wording signals are reliable once looked for. Average, mean and per typically point to the sampling distribution of the mean; total, combined, altogether and sum point to the sum; a randomly selected one, a single, or this customer point to the population itself. When none of these appears, the safest move is to ask what units the answer should carry and how many observations it summarises.
Figure (svg): A decision procedure for choosing between the population, the mean and the sum
OpenStax Introductory Statistics 2e, §7.3 Using the Central Limit Theorem §7.3, pp. 373-374 — the instruction and the note about individual values
Picture it
The book's own decision rule, with its warning attached.
Figure (svg): A decision procedure for choosing between the population, the mean and the sum
Notice that the last line extends the rule to percentiles. It would be easy to think of the choice as belonging only to probability questions, but a percentile of the sample mean and a percentile of the population are as different as their probabilities, and section 7.1 showed the gap can be over twenty years.
Worked example
Example 7.8, whose four parts need three different treatments.
\[ \text{stress scores} \sim U(1, 5), \; n = 75 \]
Mean stress score below 2
Why: An average of 75.
90th percentile of the mean
Why: Still an average.
Total below 200
Why: A sum of 75.
90th percentile of the total
Why: Still a sum.
Figure (svg): The solution to Worked example sorting the parts of one problem shown as a ladder of expressions, one row per legal move
\[ \bar{X} \sim N(3, 0.1333), \qquad \sum X \sim N(225, 10) \]
Verify: confirm the two distributions are consistent with each other
Why: The sum's centre of 225 is 75 times the mean's centre of 3, and its spread of 10 is 75 times the mean's 0.1333 — so the two describe the same sample seen at two scales, exactly as section 7.2's conversion requires. If that factor of 75 did not hold between them, one of the two would have been computed wrongly.
OpenStax Introductory Statistics 2e, §7.3 Using the Central Limit Theorem §7.3, pp. 374-376
Sorting
Read what quantity the question is about.
Sort into buckets
Sort each question for a population with a stated distribution.
Items (a) and (e) both concern single observations, and for a skewed population their answers can differ enormously from the corresponding sampling-distribution answers. That is the section's central warning, put as a sorting task.
Worked example
Example 7.9(b), where the theorem must be set aside.
\[ X \sim \text{Exp}\left(\tfrac{1}{22}\right); \text{ find } P(x > 20) \text{ for ONE customer} \]
Read the question
Why: One randomly selected customer.
Reject the theorem
Why: It says nothing about individuals.
Use the population
Why: The exponential right tail.
\[ e ^{-\frac{20}{22}} \]
Evaluate
Why: The book's value.
\[ 0.4029 \]
Figure (svg): The solution to Worked example a part that needs the population shown as a ladder of expressions, one row per legal move
\[ P(x > 20) = e^{-20/22} \approx 0.4029 \]
Verify: confirm how wrong the theorem would have been here
Why: Treating the individual as normal with mean 22 and spread 22 would give about 0.536, and using the sampling distribution would give 0.7919 — neither is close to the correct 0.4029. The exponential's skew is exactly why: it puts far more mass at small values than any symmetric curve would, so the right tail is thinner than a normal's.
OpenStax Introductory Statistics 2e, §7.3 Using the Central Limit Theorem §7.3, p. 377
Trap
\[ P(x > 20) \text{ computed on } N(22, 2.46) \]
Use the sampling distribution because the chapter is about it
Why: Two sections have just established that things become normal.
\[ 0.7919 \text{ instead of } 0.4029 \]
The 2.46 describes how MEANS of eighty customers vary, and one customer varies with a standard deviation of 22.
\[ P(x > 20) = e^{-20/22} \approx 0.4029, \text{ from the exponential itself} \]
Use the random variable's own distribution for an individual
Why: The theorem makes no claim about single observations.
The book states this as a standalone note for good reason: it is the error the chapter itself sets up. Two sections of normal approximations create a habit, and this question breaks it. The defence is to name the quantity — one customer, or a mean of eighty — before choosing any distribution.
Fill the middle
What to do when the question asks about one observation.
Fill in the blanks
\textrandom variable ___
Why: Its own random variable — the exponential, the uniform, the binomial, whatever the problem states. The central limit theorem is about means and sums only.
Two truths and a lie
All three concern choosing.
Eliminate the wrong options
Two are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is false and misreads the theorem entirely. Individual values keep whatever distribution they always had, however large the sample; it is the MEANS and SUMS that become normal. A population of exponential waiting times stays exponential no matter how many are sampled.
Prediction
Commit before reasoning.
Predict first
Which phrase signals that the sum's distribution is needed?
Correct: The combined total of the sample.
Why: Total, combined, altogether and sum all point to the sum's distribution. Average and mean point to the sampling distribution of the mean, and a randomly selected individual points to the population's own. The fourth option names a parameter rather than a random quantity, so no distribution is needed for it at all.
Section
Section 2
Concept
The law of large numbers says that if you take samples of larger and larger size from any population, then the mean of the sample tends to get closer and closer to mu. The book adds that the central limit theorem illustrates it, because the standard error gets smaller as n grows.
the law of large numbers — The statement that mu is the value the sample means approach as n gets larger. It follows from the standard error, sigma over root n, shrinking toward zero.
\[ \frac{\sigma}{\sqrt{n}} \to 0 \;\text{ as }\; n \to \infty \]
The two results are worth distinguishing. The law of large numbers says where the sample means end up; the central limit theorem says how they are scattered on the way there. The second is the stronger statement, and it implies the first — which is exactly the book's remark that the theorem illustrates the law.
Figure (svg): A horizontal dashed line for the population mean with a funnel of two curves narrowing toward it as the sample size grows
OpenStax Introductory Statistics 2e, §7.3 Using the Central Limit Theorem §7.3, p. 374 — the law of large numbers
Picture it
Two standard errors either side of mu, as the sample size grows.
Figure (svg): A horizontal dashed line for the population mean with a funnel of two curves narrowing toward it as the sample size grows
The funnel narrows quickly at first and then very slowly, which is the square root again. It never closes completely for any finite n, which is why a sample mean is always an estimate with uncertainty attached rather than the population mean itself — and quantifying that uncertainty is what chapter 8 is about.
Worked example
Two standard errors for a population with sigma equal to 15.
\[ \sigma = 15; \; n = 10, 40, 160, 640 \]
At n = 10
Why: Fifteen over root 10, doubled.
\[ \text{about } 9.5 \]
At n = 40
Why: Four times the sample.
\[ \text{about } 4.7 \]
At n = 160
Why: Four times again.
\[ \text{about } 2.4 \]
At n = 640
Why: And again.
\[ \text{about } 1.2 \]
Figure (svg): The solution to Worked example how close, at four sample sizes shown as a ladder of expressions, one row per legal move
\[ 2\cdot\frac{15}{\sqrt{n}}: \; 9.5 \to 4.7 \to 2.4 \to 1.2 \]
Verify: confirm the law of large numbers holds in the limit
Why: The sequence is halving without bound, so it approaches zero — which is precisely the claim that sample means approach mu. But it never reaches zero for finite n, so no sample ever pins mu down exactly. Both halves of that statement matter: the law guarantees convergence, and the standard error quantifies how far off you still are.
OpenStax Introductory Statistics 2e, §7.3 Using the Central Limit Theorem §7.3, p. 374
Two truths and a lie
All three concern the law.
Eliminate the wrong options
Two are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is false and is the gambler's fallacy. Independent observations have no memory, so a run changes nothing about what comes next. The average returns toward mu because later observations outnumber the run, not because anything compensates for it.
Worked example
Distinguishing the law from a common misreading.
\[ \text{a fair coin has come up heads } 10 \text{ times running} \]
State the law
Why: Sample means approach mu.
\[ \text{the PROPORTION tends to } 0.5 \]
Ask what it says about the next flip
Why: Nothing at all.
\[ \text{still } 0.5 \]
Say how the average recovers
Why: Later flips outnumber the run.
Name the misreading
Why: Expecting compensation.
Figure (svg): The solution to Worked example what the law does not promise shown as a ladder of expressions, one row per legal move
\[ P(\text{heads}) = 0.5 \text{, whatever has gone before} \]
Verify: confirm the arithmetic of dilution
Why: After 10 heads, the proportion is 1.0. After a further 990 flips averaging 0.5, the overall proportion is about 0.505 — close to a half without any run of tails ever occurring. The early excess becomes negligible because it is divided by a growing n, not because it is cancelled, and understanding that is the difference between the law and the fallacy.
OpenStax Introductory Statistics 2e, §7.3 Using the Central Limit Theorem §7.3, p. 374
Trap
\[ \text{ten heads running} \;\Rightarrow\; \text{tails is now more likely} \]
Reason that the average must return to a half
Why: The law of large numbers guarantees it.
\[ \text{but each flip is independent, so } P = 0.5 \]
The law describes a limit over many observations; it grants no influence over any particular next one.
\[ \text{the run is DILUTED, not cancelled} \]
Note that the excess is divided by a growing n rather than reversed
Why: Later data outnumbers it.
This is the same memorylessness section 5.3 met, in a different costume: the past carries no information about the next independent trial. The law of large numbers is one of the most frequently misquoted results in statistics precisely because its correct statement — about a limit — sounds so much like the incorrect one about compensation.
Faded example
A population has sigma = 15. Compare two standard errors at n = 40 and n = 160.
Fill in the blanks
\text2.4 n = 40: \; 2\cdot\frac2___} \approx 4.7; \quad \text___ n = 160: \approx ___, \text___ ___
Why: Quadrupling n halves the standard error, so 4.7 becomes about 2.4. The funnel narrows steadily but ever more slowly, which is the square root at work.
Prediction
Commit before reasoning.
Predict first
For any finite sample size, does the standard error ever reach zero?
Correct: No: it approaches zero but never reaches it.
Why: Sigma over root n is positive for every finite n, so some uncertainty always remains. That is why chapter 8 reports an interval rather than a single value, and why a poll of a thousand still carries a margin of error. The second option confuses the law with the rule of thumb about when normality takes hold.
Explain it
A classmate says the law of large numbers means a coin that has shown ten heads owes you some tails.
Discussion prompt
In two sentences or fewer, correct them.
Hint: Ask what happens to the proportion after another thousand flips.
Answer:
Each flip is independent, so the next one is still a half — the coin has no memory of the run and nothing is owed.
The proportion returns toward a half because another thousand flips swamp the ten, diluting the excess rather than cancelling it.
Section
Section 3
Concept
Example 7.8 takes stress scores that follow a uniform distribution between 1 and 5. Nothing about the population is bell shaped, yet the theorem gives the sample means a normal distribution, with the uniform's own mean and standard deviation feeding into the usual formulas.
deriving mu and sigma first — For a non-normal population, the population's mean and standard deviation must be computed from its own formulas before the theorem can be applied. For a uniform they are the midpoint and the width over the square root of twelve.
\[ \mu = \frac{a+b}{2} = 3, \qquad \sigma = \frac{b-a}{\sqrt{12}} \approx 1.15 \]
This is the first example in the chapter where the theorem is doing something chapter 6 could not. In Example 7.2 the population was already normal, so the sampling distribution's normality was no news; here the population is flat, and the bell shape of its sample means is a genuine consequence of the theorem rather than an inheritance from the population.
Figure (svg): A flat rectangle for the population with three progressively narrower bell curves above it, showing sample means becoming normal even though the population is uniform
OpenStax Introductory Statistics 2e, §7.3 Using the Central Limit Theorem §7.3, pp. 374-376 — Example 7.8
Picture it
The uniform population, and the means of 5, 25 and 75 drawn from it.
Figure (svg): A flat rectangle for the population with three progressively narrower bell curves above it, showing sample means becoming normal even though the population is uniform
The population is drawn as a rectangle because that is what it is — no peak, no tails, every score equally likely. The three curves above it are all bell shaped, and they get narrower in the ratio of one over the square root of n. Neither feature is present in the rectangle they came from.
Worked example
Example 7.8(b).
\[ X \sim U(1,5), \; n = 75; \text{ find the } 90\text{th percentile of } \bar{X} \]
Find the population's mean
Why: The midpoint of 1 and 5.
\[ 3 \]
Find its standard deviation
Why: Four over root 12.
\[ 1.15 \]
Find the standard error
Why: 1.15 over root 75.
\[ 0.1333 \]
Reverse the cumulative
Why: At 0.90.
\[ 3.2 \]
Figure (svg): The solution to Worked example the 90th percentile of the mean stress score shown as a ladder of expressions, one row per legal move
\[ k = \text{invNorm}(0.90,\; 3,\; 0.1333) \approx 3.2 \]
Verify: confirm the answer against the population's own 90th percentile
Why: For individual scores the 90th percentile of U(1, 5) is 1 plus 0.9 times 4, which is 4.6 — far above the 3.2 for a mean of 75. The gap is the concentration the theorem describes, and it is why the interpreting sentence has to say means of 75 scores. A percentile of 4.6 quoted for a sample mean would be badly wrong.
OpenStax Introductory Statistics 2e, §7.3 Using the Central Limit Theorem §7.3, p. 375
Faded example
Stress scores are uniform between 1 and 5.
Fill in the blanks
\mu = \frac31.15 = ___, \qquad \sigma = \frac______} \approx ___
Why: The uniform's mean is the midpoint, 3, and its standard deviation is the width of 4 divided by about 3.464, giving 1.15 — the book's rounded value. Both must be found before the theorem can be applied.
Worked example
Example 7.8(c) and (d).
\[ \sum X \sim N(225, 10); \text{ find } P\left(\sum x < 200\right) \text{ and the } 90\text{th percentile} \]
The sum's centre
Why: 75 times 3.
\[ 225 \]
The sum's spread
Why: Root 75 times 1.1547.
\[ 10 \]
The probability
Why: 200 is 2.5 spreads below.
\[ 0.0062 \]
The 90th percentile
Why: Reverse at 0.90.
\[ 237.8 \]
Figure (svg): The solution to Worked example the total of 75 stress scores shown as a ladder of expressions, one row per legal move
\[ P\left(\sum x < 200\right) \approx 0.0062, \qquad k \approx 237.8 \]
Verify: confirm both against the mean's scale
Why: Dividing 200 by 75 gives 2.667, and dividing 237.8 by 75 gives 3.17 — both sensible as average stress scores, and the second matches the 90th percentile of the mean found earlier. The book also notes a real-world constraint the model ignores: the smallest possible total is 75, since the smallest single score is one, so the normal curve's left tail extends into territory the data cannot reach.
OpenStax Introductory Statistics 2e, §7.3 Using the Central Limit Theorem §7.3, pp. 375-376
Error analysis
The population is U(1, 5) and n is 75; the correct standard error is about 0.1333.
Annotate
On: \( \begin{aligned} &(1)\; \frac{5}{\sqrt{75}} \approx 0.577 \\ &(2)\; \frac{1.1547}{75} \approx 0.0154 \\ &(3)\; 1.1547 \\ &(4)\; \frac{1.1547}{\sqrt{75}} \approx 0.1333 \end{aligned} \)
Error (1) is specific to non-normal populations and is the one this idea exists to prevent: the population's own parameters have to be DERIVED before the theorem can be used, and for a uniform neither of them is any of the numbers written in the problem.
Two truths and a lie
All three concern non-normal populations.
Eliminate the wrong options
Two are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is false. The population is a fixed thing and does not change with the sample size — stress scores remain uniformly distributed however many are collected. Only the distribution of their MEANS becomes normal.
Estimation
For means of 75 stress scores, the book finds P(mean < 2) is about zero.
Predict first
Why is that, when a score of 2 is perfectly ordinary?
Correct: Because 2 is about 7.5 standard errors below 3.
Why: The standard error is only 0.1333, so a full point below the mean is an enormous distance on the sampling scale — even though a quarter of individual scores fall below 2. Getting all 75 of them to average below 2 is what is essentially impossible, not any single score being low.
Prediction
Commit before reasoning.
Predict first
The normal model for the total of 75 stress scores extends below 75. What is wrong with that?
Correct: The smallest possible total is 75.
Why: The book adds this reminder explicitly. A normal distribution has infinite tails, so it assigns positive probability to totals the data could never produce — the approximation is good in the middle and imperfect at the extremes. The upper limit of 375 is also real, but the book raises the lower one because that is the tail the question asks about.
Section
Section 4
Concept
Example 7.9 takes excess mobile minutes following an exponential distribution with a mean of 22, so the population's mean and standard deviation are both 22. The mean of eighty such values is approximately normal with a standard error of 22 over the square root of eighty, and the same threshold gives very different answers on the two distributions.
the two answers — P(x > 20) is 0.4029 for one customer and P(x-bar > 20) is 0.7919 for a mean of eighty. The book's explanation is that the probabilities are not equal because we use different distributions to calculate the probability for individuals and for means.
\[ P(x > 20) = 0.4029 \quad\text{against}\quad P(\bar{x} > 20) = 0.7919 \]
The reason for the gap is worth spelling out. Twenty is below the mean of 22, so a majority of the distribution lies above it either way — but for the exponential the majority is slight, because the long right tail is offset by a great mass of very small values just above zero. For the sample mean, twenty is 0.81 standard errors below a nearly symmetric centre, so nearly four fifths of the distribution lies above it.
Figure (svg): Two panels: a declining exponential curve with its right tail beyond twenty shaded, beside a narrow bell curve with most of its area beyond twenty shaded
OpenStax Introductory Statistics 2e, §7.3 Using the Central Limit Theorem §7.3, pp. 376-378 — Example 7.9 and its explanation
Picture it
Example 7.9: twenty minutes, asked of one customer and of a mean of eighty.
Figure (svg): Two panels: a declining exponential curve with its right tail beyond twenty shaded, beside a narrow bell curve with most of its area beyond twenty shaded
The left panel's curve is the exponential of section 5.3, unchanged; the right panel's is the normal the theorem produces. Both shade the region above twenty, and the shaded fractions are visibly different — which is the picture the book's explanation describes in words.
Worked example
Example 7.9(a).
\[ X \sim \text{Exp}\left(\tfrac{1}{22}\right), \; n = 80; \text{ find } P(\bar{x} > 20) \]
Recall the population's parameters
Why: For an exponential, mu equals sigma.
\[ \text{both } 22 \]
Find the standard error
Why: 22 over root 80.
\[ 2.46 \]
Locate 20
Why: Below the centre.
\[ 0.81\text{ SEs below} \]
Take the right tail
Why: One minus the left area.
\[ 0.7919 \]
Figure (svg): The solution to Worked example the mean excess time above 20 minutes shown as a ladder of expressions, one row per legal move
\[ P(\bar{x} > 20) \approx 0.7919 \]
Verify: confirm the exponential's mean and standard deviation really do coincide
Why: Section 5.3 established that an exponential has sigma equal to mu, so a mean of 22 gives a standard deviation of 22 with nothing further to compute. That is unusual enough to be worth checking rather than assuming — for most distributions the two would have to be derived separately, as they did for the uniform in the previous idea.
OpenStax Introductory Statistics 2e, §7.3 Using the Central Limit Theorem §7.3, p. 377
Faded example
Excess minutes are exponential with mean 22, so m is 1/22.
Fill in the blanks
P(x > 20) = e^0.4029 \approx exponential, \text___ ___ \text___
Why: The exponential's right tail is e to the minus m x, giving 0.4029. Using the sampling distribution here would give 0.7919, which answers a completely different question.
Worked example
Example 7.9's percentile part.
\[ \bar{X} \sim N(22, 2.46); \text{ find the } 95\text{th percentile} \]
Set the area
Why: Ninety-five percent below.
\[ 0.95 \]
Recall the z-score
Why: For a 95th percentile.
\[ 1.645 \]
Scale and shift
Why: 22 plus 1.645 times 2.46.
\[ 26.0 \]
Interpret
Why: The book's own sentence.
\[ \text{samples of } 80 \]
Figure (svg): The solution to Worked example the 95th percentile of the sample mean shown as a ladder of expressions, one row per legal move
\[ k \approx 26.0 \text{ minutes} \]
Verify: confirm against the population's own 95th percentile
Why: For an individual, the 95th percentile of an exponential with mean 22 is 22 times the natural log of 20, about 66 minutes — two and a half times the sample mean's 26. The book's interpretation is careful to say samples rather than customers for exactly this reason, and it adds that only five percent of such samples would have means above 26 minutes.
OpenStax Introductory Statistics 2e, §7.3 Using the Central Limit Theorem §7.3, pp. 377-378
Trap
\[ 0.4029 \text{ and } 0.7919 \text{ cannot both be right} \]
Expect an individual and an average to behave alike
Why: They come from the same population, after all.
\[ \text{but they are different random quantities} \]
The population is strongly skewed while the mean of eighty is nearly symmetric, so a threshold below the centre sits very differently on the two.
\[ \text{different distributions, therefore different probabilities} \]
Accept that a mean is a different quantity from an observation
Why: The book's own resolution: we use different distributions for individuals and for means.
The gap is largest exactly where the population is most skewed, which is why the book chose an exponential to make the point. For a symmetric population the two answers would still differ — because the spreads differ — but not so dramatically, and the lesson would be easier to miss.
Two truths and a lie
All three concern Example 7.9.
Eliminate the wrong options
Two are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is false. Both answers are correct for the questions they answer: 0.4029 is about one customer and 0.7919 is about a mean of eighty. They are different quantities, so there is no contradiction to resolve.
Estimation
Compare a skewed population with a symmetric one.
Predict first
For which population is the gap between P(x > c) and P(x-bar > c) generally larger?
Correct: The strongly skewed one.
Why: For a skewed population the individual distribution differs from a normal in shape as well as spread, so both effects push the two answers apart. For a symmetric population only the spread differs. That is why the book chose an exponential to illustrate the point rather than something closer to normal.
Explain it
A classmate cannot see how the same threshold gives 0.4029 and 0.7919 for the same customers.
Discussion prompt
In three sentences or fewer, resolve it.
Hint: Ask what quantity each probability is about.
Answer:
The first is about one customer's own excess time, which follows a strongly skewed exponential with a spread of 22 minutes.
The second is about the AVERAGE of eighty customers, which is nearly symmetric with a spread of only 2.46 minutes.
Twenty minutes is barely below the centre either way, but on the tight sampling distribution almost everything sits above it, while on the skewed population a great mass of very small values sits below.
Section
Section 5
Concept
Every percentile in this chapter belongs to one of three distributions, and its interpretation has to say which. The book's own sentences are careful about this throughout, and the gap between the readings can be very large.
interpreting in context — Naming the quantity, the sample size and the direction: not 26.0 but ninety-five percent of samples of 80 customers would have mean excess times under 26 minutes.
\[ 26.0 \;\longrightarrow\; \text{“95 percent of SAMPLES of 80 have means under 26 minutes”} \]
The scale check is the practical companion to the sentence. A percentile of a sum should be roughly n times a plausible individual value, a percentile of a mean should be close to mu, and a percentile of the population should look like an ordinary observation. An answer that fails that rough test has almost always been computed on the wrong distribution.
Figure (svg): A normal curve for the total of seventy-five stress scores with the far left tail below two hundred shaded
OpenStax Introductory Statistics 2e, §7.3 Using the Central Limit Theorem §7.3, pp. 375-378 — the interpretations in Examples 7.8 and 7.9
Picture it
Example 7.8(c): the chance seventy-five stress scores total below 200.
Figure (svg): A normal curve for the total of seventy-five stress scores with the far left tail below two hundred shaded
The answer of 0.0062 is small, and the picture agrees. The book's reminder that the smallest possible total is 75 is worth carrying: the normal model's left tail runs off toward values that cannot occur, which is a limitation of the approximation rather than a fact about stress scores.
Worked example
Comparing the readings across the chapter's three distributions.
\[ \text{stress scores } U(1,5), \; n = 75 \]
For one score
Why: The uniform's own.
\[ 4.6 \]
For the mean of 75
Why: N(3, 0.1333).
\[ 3.2 \]
For the total of 75
Why: N(225, 10).
\[ 237.8 \]
Check the last two agree
Why: 237.8 over 75.
\[ 3.17 \]
Figure (svg): The solution to Worked example three percentiles, three sentences shown as a ladder of expressions, one row per legal move
\[ 4.6, \quad 3.2, \quad 237.8 \]
Verify: confirm the mean and sum percentiles are consistent
Why: Dividing the total's percentile of 237.8 by 75 gives 3.17, which matches the mean's percentile of about 3.2 to rounding — as it must, since a total below k is the same event as a mean below k over 75. The individual's 4.6 has no such relationship to either, because it describes a different kind of quantity altogether.
OpenStax Introductory Statistics 2e, §7.3 Using the Central Limit Theorem §7.3, pp. 374-376
Discrimination
Stress scores run from 1 to 5, and n is 75.
Sort into buckets
Sort each candidate answer by whether its scale is plausible.
Worked example
Example 7.9's percentile, in the book's own words.
\[ k = 26.0 \text{ for } \bar{X} \sim N(22, 2.46) \]
Name the quantity
Why: A sample mean.
Name the sample size
Why: Eighty customers.
\[ \text{samples of } 80 \]
Give the direction
Why: Ninety-five percent below.
\[ \text{under } 26\text{ minutes} \]
Add the complement
Why: As the book does.
\[ 5 \%\text{ above} \]
Figure (svg): The solution to Worked example writing the interpretation shown as a ladder of expressions, one row per legal move
\[ P(\bar{X} < 26.0) = 0.95 \]
Verify: confirm the sentence would be false if it named customers instead
Why: Ninety-five percent of individual customers do not have excess times under 26 minutes — the exponential's 95th percentile is about 66 minutes, so roughly a quarter of customers exceed 26. The sentence is not merely imprecise if it names customers; it is wrong by a wide margin, which is why the book writes samples explicitly.
OpenStax Introductory Statistics 2e, §7.3 Using the Central Limit Theorem §7.3, pp. 377-378
Trap
\[ \text{the } 90\text{th percentile of the total} \approx 3.2 \]
Use the mean's distribution for a question about a total
Why: Both are about samples of 75.
\[ 3.2 \text{ as a total of seventy-five scores} \]
Seventy-five scores each at least 1 must total at least 75, so a total of 3.2 is impossible.
\[ \text{the } 90\text{th percentile of the total} \approx 237.8 \]
Match the distribution to the quantity, then sanity-check the scale
Why: A total of 75 scores belongs in the hundreds.
The scale check is what makes this error self-announcing, and it is worth applying to every answer in the chapter. A total should be roughly n times an individual value, a mean should be near mu, and an individual should look like an ordinary observation — three quick tests that between them catch nearly every misapplied distribution.
Two truths and a lie
All three concern interpretation.
Eliminate the wrong options
Two are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is false. For Example 7.8 they are 3.2 and 4.6, and for Example 7.9 about 26 and 66. They coincide only when n equals 1, since only then is the sampling distribution the population's own.
Faded example
The 90th percentile of the total of 75 stress scores is 237.8.
Fill in the blanks
\frac753.17} \approx ___, \text___ 90\text___
Why: Dividing by 75 gives about 3.17, matching the 3.2 found on the mean's distribution. The two must agree because a total below k is the same event as a mean below k over 75.
Prediction
Commit before reasoning.
Predict first
A student reports the 90th percentile of a total of 75 scores as 3.2. What most likely went wrong?
Correct: The mean's distribution was used instead of the sum's.
Why: 3.2 is precisely the mean's 90th percentile, so the right calculation was performed on the wrong distribution. The scale test spots it instantly — a total of 75 scores each at least 1 cannot be below 75 — without needing to know what the correct answer is.
Comparison
Fill the blanks. The whole of chapter 7 on one table.
Comparison matrix
| Question is about | Distribution | Centre and spread |
|---|---|---|
| One individual value | the population's own | mu and sigma |
| A mean of n | normal, by the theorem | mu and sigma over root n |
| A sum of n | normal, by the theorem | n mu and root n sigma |
| Example 7.9's spread | 22 for one customer | 2.46 for a mean of 80 |
The first row is the one the chapter makes it easy to forget. Two sections of normal approximations build a habit, and the book's standalone warning exists because that habit is wrong exactly when the question turns to a single observation.
Pattern
Six steps, and the first is the one this whole section is about.
Two consistency checks are free: a sum's answer divided by n should match the mean's answer, and a total of n values must lie within n times the population's own range.
OpenStax Introductory Business Statistics 2e, §7.2 Using the Central Limit Theorem §7.2 Using the Central Limit Theorem
Check
Choosing the distribution.
Check your understanding
Excess minutes are exponential with mean 22. Which distribution answers: what is the probability that one randomly chosen customer exceeds 20 minutes?
Answer: A
Why: The question is about one individual value, so the central limit theorem does not apply and the variable's own distribution is used.
Check
A non-normal population.
Check your understanding
Stress scores are uniform on 1 to 5. What is the population's standard deviation?
Answer: A
Why: For a uniform the standard deviation is the width divided by the square root of twelve: 4 over about 3.464, which is 1.15.
Check
The law of large numbers.
Check your understanding
A fair coin shows ten heads in a row. What does the law of large numbers predict about the next flip?
Answer: A
Why: The law describes what happens to the average over many flips, not to any particular next one. Independent flips have no memory.
Real world
A hospital's emergency department reports that individual patient waiting times are strongly right-skewed, with a mean of 45 minutes and a standard deviation of 45 minutes. A regulator sets a target that the average wait across each day's 100 patients must not exceed 50 minutes, and a manager objects that since a great many individual patients wait over 50 minutes, the target is unachievable.
Discussion prompt
Compute how often the target is missed when the process is on target, assess the manager's objection, and say what the target actually measures.
Hint: The two claims are about different quantities.
Answer:
The manager is right about individuals and wrong about the target. With a mean and standard deviation both equal to 45, the waits look exponential, so the proportion of individual patients waiting over 50 minutes is about e to the minus fifty over forty-five, roughly 0.33 — a third of all patients, which is a great many.
\[ \text{SE} = \frac{45}{\sqrt{100}} = 4.5, \qquad P(\bar{x} > 50) = P(z > 1.11) \approx 0.133 \]
But the target is about the daily AVERAGE of 100 patients, whose standard error is only 4.5 minutes. Fifty minutes is 1.11 standard errors above the mean of 45, so the target is missed on about 13 percent of days — roughly one day a week, which is demanding but plainly achievable.
The target therefore measures the department's central tendency, not any individual's experience. That is a real limitation worth stating: a department could meet the average target every day while a third of patients wait over an hour, because the average is insensitive to the shape of the tail. If the regulator's concern is the patients who wait longest, an average is the wrong instrument and a percentile target — no more than 5 percent waiting over two hours, say — would measure it directly.
Two further points belong in a careful answer. The central limit theorem is doing real work here, since the population is strongly skewed and only the averaging makes a normal calculation legitimate. And the 100 patients must be reasonably independent for the standard error to be 4.5 — on a day when the department is overwhelmed, waits are correlated, the effective spread is larger, and the 13 percent figure understates how often the target is missed.
Commit first
Answer, then rate your confidence honestly.
Predict first
When should the central limit theorem NOT be used?
Correct: When the question asks about a single individual value.
\[ \text{one value} \to \text{use } X \text{ itself}, \qquad \text{a mean or sum} \to \text{use the theorem} \]
Why: The book states this as a standalone note. A non-normal population is exactly where the theorem is most useful, so the first option inverts its purpose; a large sample and a known sigma are both conditions that help rather than hinder. Only a question about one observation falls outside its scope, because the theorem makes no claim about individual values at all.
Explain it
They computed the probability that one customer's excess time exceeds 20 minutes as 0.7919, using N(22, 2.46).
Discussion prompt
In two sentences or fewer, locate the error.
Hint: Ask what the 2.46 describes.
Answer:
Their 2.46 is the standard error, which describes how the AVERAGE of eighty customers varies — one customer varies with a standard deviation of 22, nearly ten times larger.
For a single customer the exponential applies directly, giving e to the minus twenty over twenty-two, about 0.4029.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For the first, read what quantity the question is about before touching a formula. For the second, the law is about a limit, and independent trials have no memory. For the third, a uniform's parameters are the midpoint and the width over root twelve, and an exponential's are both equal to the mean. For the fourth, name the sample size in the sentence. Do five problems of your chosen kind rather than twenty mixed ones.
Connect it up
Paper. Twenty-five minutes. This one closes the chapter, so make it a decision map.
Draw it
At the top, draw a three-way branch headed by the question what is this about, with limbs for one value, a mean of n, and a total of n — and write the distribution under each, including the warning that the first limb does not use the theorem at all. Below that, work Example 7.8 completely: derive the uniform's mean of 3 and standard deviation of 1.15 from its own formulas, then write both sampling distributions, N(3, 0.1333) for the mean and N(225, 10) for the total. Compute the 90th percentile on each, and check that the total's 237.8 divided by 75 matches the mean's 3.2. Beside it, draw the uniform population as a rectangle with three progressively narrower bell curves above it, to record that a flat population has bell-shaped means. In the middle of the page, draw Example 7.9's two panels side by side: a declining exponential with its tail above 20 shaded and labelled 0.4029, and a narrow bell centred at 22 with its region above 20 shaded and labelled 0.7919 — and write one sentence on why both are correct. At the bottom, draw the law of large numbers as a funnel of two standard errors narrowing toward mu, and write beside it the one-sentence statement of the law plus the one-sentence statement of what it does NOT promise about the next flip of a coin.
Check the middle panels by confirming the two shaded fractions really do look as different as 0.40 and 0.79 — if they look similar, the sampling distribution has been drawn too wide. Check the top branch by asking, for each of Example 7.8's four parts and Example 7.9's two, which limb it belongs to; you should get two means, two sums, one mean and one individual.
Recap
Five things, and the first must happen before any of the others.
| If you see | Then |
|---|---|
| One randomly selected, a single, this customer | Use the population's own distribution |
| Average, mean, per | Use N(mu, sigma over root n) |
| Total, combined, altogether | Use N(n mu, root n sigma) |
| A uniform population | mu is the midpoint; sigma is the width over root twelve |
| An exponential population | mu and sigma are both the stated mean |
| A run of high values | The law promises nothing about the next one |
| An answer on an impossible scale | The wrong distribution was used |
That closes chapter 7, and with it the descriptive and probabilistic half of the course. Chapter 8 turns the standard error around: instead of asking how far a sample mean is likely to fall from a known mu, it asks what values of mu are consistent with an observed sample mean — which is a confidence interval, and the first genuine act of inference in the book.
OpenStax Introductory Statistics 2e, §7.3 Using the Central Limit Theorem §7.3, pp. 373-380 — everything on these slides traces back here
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